In ΔLMN, n = 960 cm, ∠L=160° and ∠M=10°. Find the length of l, to the nearest centimeter.
2 answers:
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Answer:
a) 40.17% probability of a value between 75.0 and 90.0.
b) 35.94% probability of a value 75.0 or less.
c) 20.22% probability of a value between 55.0 and 70.0.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a. Compute the probability of a value between 75.0 and 90.0.
This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 75.
X = 90
has a pvalue of 0.7611
X = 75
has a pvalue of 0.3594
0.7611 - 0.3594 = 0.4017
40.17% probability of a value between 75.0 and 90.0.
b. Compute the probability of a value 75.0 or less.
This is the pvalue of Z when X = 75. So
has a pvalue of 0.3594
35.94% probability of a value 75.0 or less.
c. Compute the probability of a value between 55.0 and 70.0.
This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 55.
X = 70
has a pvalue of 0.2389
X = 55
has a pvalue of 0.0367
0.2389 - 0.0367 = 0.2022
20.22% probability of a value between 55.0 and 70.0.